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Proof step 6 from proof byte-mult-pos (previous step)

¬(0 ≤ i * 256 + j ∧ 127 < (i * 256 + j) % 256), 0 ≤ i, 0 ≤ j ⊦ b2i(i2b(i * 256 + j)) = b2i(i2b(j))

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

name: bib-pos-low rotate: F subst: [i] → [i * 256 + j]

Additional information

Used simplifier rules: byte-equal: ⊦ b2i(byi) = b2i(byj) ↔ byi = byj;%bib-pos-low: ⊦ 0 ≤ i ∧ i % 256 ≤ 127 → b2i(i2b(i)) = i % 256;% ⊦ 0 ≤ j ∧ 0 ≤ j0 ∧ 0 ≤ i0 → 0 ≤ i0 * j + j0;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ 0 < j ∧ 0 ≤ i ∧ 0 ≤ i0 → (j * i + i0) % j = i0 % j;% ⊦ ¬ i < j ↔ j ≤ i;%j0 ≠ 0, j ≤ j0 ⊦ i ≤ (* j) → i ≤ (* j0); Used terms: i * 256 + j